Django models - how to filter number of ForeignKey objects

Question

I have a models A and B, that are like this:

class A(models.Model):
  title = models.CharField(max_length=20)
  (...)

class B(mode         


        

Answer 1

I assume that joining or leaving the pool may not happen as often as listing (showing) the pools. I also believe that it would be more efficient for the users join/leave actions to update the pool display status. This way, listing & showing the pools would require less time as you would just run a single query for SHOW_STATUS of the pool objects.

Answer 2

The question and selected answer are from 2008 and since then this functionality has been integrated into the django framework. Since this is a top google hit for "django filter foreign key count" I'd like to add an easier solution with a recent django version using Aggregation.

from django.db.models import Count
cats = A.objects.annotate(num_b=Count('b')).filter(num_b__lt=2)

In my case I had to take this concept a step further. My "B" object had a boolean field called is_available, and I only wanted to return A objects who had more than 0 B objects with is_available set to True.

A.objects.filter(B__is_available=True).annotate(num_b=Count('b')).filter(num_b__gt=0).order_by('-num_items')

Answer 3

I'd recommend modifying your design to include some status field on A.

The issue is one of "why?" Why does A have < 2 B's and why does A have >= 2 B's. Is it because user's didn't enter something? Or is because they tried and their input had errors. Or is it because the < 2 rule doesn't apply in this case.

Using presence or absence of a Foreign Key limits the meaning to -- well -- present or absent. You don't have any way to represent "why?"

Also, you have the following option

[ a for a in A.objects.all() if a.b_set.count() < 2 ]

This can be pricey because it does fetch all the A's rather than force the database to do the work.

---

Edit: From the comment "would require me to watch for user join / user leaving the pool events".

You don't "watch" anything -- you provide an API which does what you need. That's the central benefit of the Django model. Here's one way, with explict methods in the A class.

class A( models.Model ):
    ....
    def addB( self, b ):
        self.b_set.add( b )
        self.changeFlags()
    def removeB( self, b ):
        self.b_set.remove( b )
        self.changeFlags()
    def changeFlags( self ):
        if self.b_set.count() < 2: self.show= NotYet
        else: self.show= ShowNow

You can also define a special Manager for this, and replace the default b_set Manager with your manager that counts references and updates A.

Answer 4

Sounds like a job for extra.

A.objects.extra(
    select={
        'b_count': 'SELECT COUNT(*) FROM yourapp_b WHERE yourapp_b.a_id = yourapp_a.id',
    },
    where=['b_count < 2']
)

If the B count is something you often need as a filtering or ordering criterion, or needs to be displayed on list views, you could consider denormalisation by adding a b_count field to your A model and using signals to update it when a B is added or deleted:

from django.db import connection, transaction
from django.db.models.signals import post_delete, post_save

def update_b_count(instance, **kwargs):
    """
    Updates the B count for the A related to the given B.
    """
    if not kwargs.get('created', True) or kwargs.get('raw', False):
        return
    cursor = connection.cursor()
    cursor.execute(
        'UPDATE yourapp_a SET b_count = ('
            'SELECT COUNT(*) FROM yourapp_b '
            'WHERE yourapp_b.a_id = yourapp_a.id'
        ') '
        'WHERE id = %s', [instance.a_id])
    transaction.commit_unless_managed()

post_save.connect(update_b_count, sender=B)
post_delete.connect(update_b_count, sender=B)

Another solution would be to manage a status flag on the A object when you're adding or removing a related B.

B.objects.create(a=some_a)
if some_a.hidden and some_a.b_set.count() > 1:
    A.objects.filter(id=some_a.id).update(hidden=False)

...

some_a = b.a
some_b.delete()
if not some_a.hidden and some_a.b_set.count() < 2:
    A.objects.filter(id=some_a.id).update(hidden=True)